# Connected and Disconnected Metric Spaces

Definition: A metric space $(M, d)$ is said to be Disconnected if there exists nonempty open sets $A$ and $B$ such that $A \cap B = \emptyset$ and $M = A \cup B$. If $M$ is not disconnected then we say that $M$ Connected. Furthermore, if $S \subseteq M$ then $S$ is said to be disconnected/connected if the metric subspace $(S, d)$ is disconnected/connected. |

Intuitively, a set is disconnected if it can be separated into two pieces while a set is connected if itâ€™s an entire piece.

For example, consider the metric space $(\mathbb{R}, d)$ where $d$ is the Euclidean metric on $\mathbb{R}$. Let $S = (a, b) \subset \mathbb{R}$, i.e., $S$ is an open interval in $\mathbb{R}$. We claim that $S$ is connected.

Suppose not. Then there exists nonempty open subsets $A$ and $B$ such that $A \cap B = \emptyset$ and $(a, b) = A \cup B$. Furthermore, $A$ and $B$ must be open intervals themselves, say $A = (c, d)$ and $B = (e, f)$. We must have that $A \cup B = (c, d) \cup (e, f)$. So $c = a$ or $e = a$ and furthermore, $d = b$ or $f = b$.

If $c = a$ then this implies that $f = b$ (since if $d = b$ then $A = (a, b)$ which implies that $B = \emptyset$). So if $A \cup B = (c, d) \cup (e, f) = (a, d) \cup (e, b) $ we must have that [[$ a < d, e < b$. If $d = e$ then $A \cup B = (a, d) \cup (d, b)$ and so $d \not \in (a, b)$ so $A \cup B \neq = (a, b)$. If $d < e$ then $A \cup B = (a, d) \cup (e, b)$ and $(d, e) \not \in (a, b)$ so $A \cup B \neq (a, b)$. If $d > e$ then $A \cap B = (e, d) \neq \emptyset$. Either way we see that $(a, b) \neq A \cup B$.

We can use the same logic for the other cases which will completely show that $(a, b)$ is connected.